A point charge is moving with a uniform velocity $3\hat{\mathrm{i}}$ m/s is released in an electric field of strength $\overrightarrow{\mathrm{E}}=\mathrm{b}\hat{\mathrm{j}} \mathrm{V}/\mathrm{m}.$ The subsequent trajectory of the particle will be

$\mathrm{displacement} \mathrm{in} \mathrm{x} \mathrm{direction}$

$x=3t$

$\mathrm{displacement} \mathrm{along} \mathrm{y} \mathrm{direction}$

$y= \frac{1}{2}\frac{bq}{m}{\left(t\right)}^2$

$\mathrm{substiituting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{t}$

$\mathrm{we} \mathrm{get} \mathrm{y}=\frac{1}{2}\frac{\mathrm{bq}}{\mathrm{m}}{\left(\frac{\mathrm{x}}{3}\right)}^2$

The above equation represents the equation of a parabola.